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In the book Quantum Field Theory and the Standard Model when computing the vacuum polarization diagrams in scalar QED and after some pages in QED the author uses some handwavyng arguments to compute some integrals that I really can't understand.

The first one is this:

Adding the diagrams gives

$$i\Pi^{\mu\nu}_2=-e^2\int \dfrac{d^4k }{(2\pi)^4}\dfrac{-4k^\mu k^\nu+2p^\mu k^\nu+2p^\nu k^\mu-p^\mu p^\nu+2g^{\mu\nu}[(p-k)^2-m^2]}{[(p-k)^2-m^2+i\epsilon][k^2-m^2+i\epsilon]}$$

Fortunately, we do not need to evaluate the entire integral. By looking at what possible form it could have, we can isolate the part that will contribute to a correction to Coulomb's law and just calculate that part. By Lorentz invariance, the most general form that $\Pi^{\mu\nu}_2$ could have is

$$\Pi^{\mu\nu}_2=\Delta_1(p^2,m^2)p^2 g^{\mu\nu}+\Delta_2(p^2,m^2)p^\mu p^\nu$$

After arguing the relation between $\Pi^{\mu\nu}_2$ and the dressed propagator $G^{\mu\nu}$ the author writes it as (in Feynman gauge)

$$iG^{\mu\nu}(p)=\dfrac{(1+\Delta_1)g^{\mu\nu}+\Delta_2 \frac{p^\mu p^\nu}{p^2}}{p^2+i\epsilon}$$

He then says that:

Next note that the $\Delta_2$ term is just a change of gauge - it gives a correction to the unphysical gauge parameter $\xi$ in covariant gauges. Since $\xi$ drops out of any physical process, by gauge invariance, so will $\Delta_2$. Thus we only need to compute $\Delta_1$. This means extracting the term proportional to $g^{\mu\nu}$ in $\Pi^{\mu\nu}$.

Most of the terms in the amplitude in Eq. (16.24) cannot give $g^{\mu\nu}$. For example, the $p^\mu p^\nu$ term must be proportional to $p^\mu p^\nu$ and can therefore only contribute to $\Delta_2$, so we can ignore it. For the $p^\mu k^\nu$ term, we can pull $p^\mu$ out of the integral, so whatever the remaining integral gives, it must provide a $p^\nu$ by Lorentz invariance. So these terms can be ignored too. The $k^\mu k^\nu$ term is important - it may give a $p^\mu p^\nu$ piece, but may also give a $g^{\mu\nu}$ piece, which is what we are looking for. So we only need to consider

$$\Pi^{\mu\nu}_2=ie^2\int\dfrac{d^4k}{(2\pi)^4}\dfrac{-4k^\mu k^\nu+2g^{\mu\nu}[(p-k)^2-m^2]}{[(p-k)^2-m^2+i\epsilon][k^2-m^2+i\epsilon]}$$

I have three issues here:

  1. First I can't understand why $\Pi^{\mu\nu}_2$ must have that form. Since $p$ appears in the integral as a parameter, I do understand that $\Pi^{\mu\nu}_2$ is a function of $p$, but why it must have that form and what it has to do with Lorentz invariance, I simply can't get. The argument that "since the only available thing is $p^\mu$ the only two tensors with tho indices we can build are $g^{\mu\nu}$ and$p^\mu p^\nu$" seems handwavying to me.

  2. Second I can't understand why based on gauge invariance the author drops the $\Delta_2$ term. I mean he is just settinga bunch of integrals to zero, this doesn't make much sense to me. I mean, he is taking all terms that contribute to $\Delta_2$ and making each of them be zero individually. Why couldn't just their sum be zero? What is the point here?

  3. Finaly, I can't undersand his analysis of what terms can be set to zero based on throwing $\Delta_2$ out. I know that if we factor a $p^\mu p^\nu$ the integral will be a scalar multiplying it, but why the integral with one $p^\mu$ factored out must by Lorentz invariance give something proportional to $p ^\nu$?

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  • $\begingroup$ As for 1., you may find 212856 useful. $\endgroup$ Commented Jul 4, 2017 at 17:57
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    $\begingroup$ As for 2., note that Schwartz talks in some depth earlier in the book about the fact that $p^\mu p^\nu$ terms in the propagator cannot modify physical observables, in chapter 8 and section 8.5 specifically. $\endgroup$
    – gj255
    Commented Jul 4, 2017 at 18:37

1 Answer 1

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  1. Given $g^{\mu\nu}$ and $p^{\mu}$, there simply are only two tensors of rank 2 that you can produce, namely $g^{\mu\nu}$ and $p^\mu p^\nu$. The only operations that alter the rank are contraction and the tensor product - but to get a rank 2 tensor, we should start with at least a rank 4 tensor which is some product of $g^{\mu\nu}$s and $p^\mu$s, but contracting any pair of indices there gives either $g^{\mu\nu}p_\nu = p^\mu$, i.e. just eliminates one factor of $g$, or $p^\mu p_\mu$, i.e. just gives a factor of $p^2$ and a tensor of 2 ranks lower. Therefore, the only independent tensors we can build are the two rank 2 tensors we get directly, namely $g^{\mu\nu}$ and $p^\mu p^\nu$, and a general rank 2 tensor that is a function of $g$ and $p$ is a combination $$ f(p^2)g^{\mu\nu} + h(p^2) p^\mu p^\nu.$$

  2. The author does not claim that $\Delta_2$ is zero, or any of the integrals contributing to it. They are saying that $\Delta_2$ is physically a "quantum correction" to the Feynman gauge parameter $\xi$, but since that parameter is completely arbitrary to begin with and has no physical meaning we can just ignore $\Delta_2$, regardless of what value it has.

  3. This is the same reasoning as in step 1, just for a rank 1 tensor: The integral with one $p^\mu$ removed is a rank 1 tensor function of $g^{\mu\nu}$ and $p^\mu$, but the only rank 1 tensor you can ever get from that is $p^\mu$ itself.

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  • $\begingroup$ As per the first question, I know that it is totally obvious to you but since the OP asked why it stemmed from Lorentz invariance, perhaps you could say a few works about why it's not possible to build a $\Pi^{\mu\nu}$ as $\Pi^{00}=p^0p^0$, $\Pi^{0i}=-p^0p^i$, etc $\endgroup$
    – user154997
    Commented Jul 5, 2017 at 7:39

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